Advertisements
Advertisements
Question
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
Advertisements
Solution
\[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\]
\[ \Rightarrow 3A = \begin{bmatrix}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{bmatrix} \left[\text{ Multiplying each element of A by 3 }\right]\]
\[ \Rightarrow \begin{vmatrix}3A\end{vmatrix} = \left( - 1 \right)^{1 + 1} 3\left( 36 - 0 \right) + \left( - 1 \right)^{1 + 2} 0\left( 0 - 0 \right) + \left( - 1 \right)^{1 + 3} 3\left( 0 - 0 \right) = 3\left( 36 - 0 \right) - 0\left( 0 - 0 \right) + 3\left( 0 - 0 \right) \left[\text{ Expanding along }R_1 \right]\]
\[ = 3 \times 36 = 108 . . . \left( 1 \right)\]
\[ \Rightarrow \begin{vmatrix}A\end{vmatrix} = \left( - 1 \right)^{1 + 1} 1\left( 4 - 0 \right) + \left( - 1 \right)^{1 + 2} 0\left( 0 - 0 \right) + \left( - 1 \right)^{1 + 3} 1\left( 0 - 0 \right) = 1\left( 4 - 0 \right) - 0\left( 0 - 0 \right) + 1\left( 0 - 0 \right) = 4 \left[\text{ Expanding along }R_1 \right]\]
\[ \Rightarrow 27\left| A \right| = 27 \times 4 = 108 . . . \left( 2 \right)\]
\[ \therefore \left| 3A \right| = 27 \left| A \right| \left[\text{ From eqs . (1) and (2) }\right]\]
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Evaluate the following determinant:
\[\begin{vmatrix}x & - 7 \\ x & 5x + 1\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix}\]
For what value of x the matrix A is singular?
\[ A = \begin{bmatrix}1 + x & 7 \\ 3 - x & 8\end{bmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & a & a^2 - bc \\ 1 & b & b^2 - ac \\ 1 & c & c^2 - ab\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
Solve the following determinant equation:
Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
State whether the matrix
\[\begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}\] is singular or non-singular.
If I3 denotes identity matrix of order 3 × 3, write the value of its determinant.
If the matrix \[\begin{bmatrix}5x & 2 \\ - 10 & 1\end{bmatrix}\] is singular, find the value of x.
If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
If \[A + B + C = \pi\], then the value of \[\begin{vmatrix}\sin \left( A + B + C \right) & \sin \left( A + C \right) & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left( A + B \right) & \tan \left( B + C \right) & 0\end{vmatrix}\] is equal to
Solve the following system of equations by matrix method:
x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1
Solve the following system of equations by matrix method:
3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15
Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17
If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\], find x, y and z.
The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has
Let \[X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}, A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\] . If AX = B, then X is equal to
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is
What is the nature of the given system of equations
`{:(x + 2y = 2),(2x + 3y = 3):}`
If `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
Let `θ∈(0, π/2)`. If the system of linear equations,
(1 + cos2θ)x + sin2θy + 4sin3θz = 0
cos2θx + (1 + sin2θ)y + 4sin3θz = 0
cos2θx + sin2θy + (1 + 4sin3θ)z = 0
has a non-trivial solution, then the value of θ is
______.
